Apply Inheritance - IB Biology

1. A woman with type A blood has a child with type O blood. She claims a man with blood type AB is the child's father. Why is this claim inconsistent with genetic principles?
   Answer: A man with type AB blood cannot have a child with type O blood. A child with type O blood has the genotype ii. This means the child must inherit one 'i' allele from each parent. A man with type AB blood has the genotype $I^A$ $I^B$. He does not possess an 'i' allele to pass on to his offspring. Therefore, it is genetically impossible for him to be the father of a type O child.
2. [HL] In a large, randomly mating population of squirrels in Hardy-Weinberg equilibrium, the frequency of the recessive allele for an albino coat is 0.2. What percentage of the population is expected to be heterozygous carriers of the albino allele?
   Answer: 32% The frequency of the recessive allele (q) is 0.2. According to the Hardy-Weinberg principle, p + q = 1, where p is the frequency of the dominant allele. Therefore, p = 1 - 0.2 = 0.8. The frequency of the heterozygous genotype is given by the term 2pq. So, the frequency of heterozygotes is 2 p q = 2 0.8 0.2 = 2 * 0.16 = 0.32. To express this as a percentage, multiply by 100, which gives 32%.
3. In pea plants, tall (T) is dominant to dwarf (t), and purple flowers (P) are dominant to white (p). A plant with genotype TtPp is self-pollinated. What is the probability that an offspring will be phenotypically tall with white flowers?
   Answer: 3/16 This is a dihybrid cross. We can consider each trait independently. For height, the cross is Tt x Tt, and the probability of a tall phenotype (TT or Tt) is 3/4. For flower colour, the cross is Pp x Pp, and the probability of a white phenotype (pp) is 1/4. To find the probability of both events occurring together, we use the product rule: P(tall and white) = P(tall) × P(white) = (3/4) × (1/4) = 3/16.
4. In Shorthorn cattle, the alleles for red coat (R) and white coat (W) are codominant, producing roan (RW) offspring. In snapdragons, alleles for red flowers ($C^R$) and white flowers ($C^W$) show incomplete dominance, producing pink ($C^R$ $C^W$) offspring. What is the fundamental difference in the expression of the heterozygous genotypes?
   Answer: In cattle, both alleles are expressed distinctly, while in snapdragons, the heterozygote has a blended phenotype. Codominance means both alleles are simultaneously and fully expressed in the heterozygote, resulting in a phenotype with distinct characteristics of both homozygotes (e.g., patches of red and white hair in roan cattle). Incomplete dominance means the heterozygote exhibits a phenotype that is an intermediate blend of the two homozygous phenotypes (e.g., red and white mix to make pink). The F2 phenotypic ratio for a heterozygous cross is 1:2:1 in both cases.
5. [HL] A geneticist performs a dihybrid test cross (AaBb x aabb) to determine if genes A and B are linked. Which result would provide the strongest evidence that the two genes assort independently?
   Answer: The four possible phenotypes appear in an approximate 1:1:1:1 ratio. The law of independent assortment applies to genes located on different chromosomes. For a dihybrid individual (AaBb), this leads to the formation of four different gametes (AB, Ab, aB, ab) in equal proportions. When this individual is test-crossed with a homozygous recessive individual (aabb), the phenotypic ratio of the offspring directly reflects the gametic ratio of the heterozygous parent. Therefore, observing the four possible phenotypes in a 1:1:1:1 ratio is the expected outcome for independent assortment.
6. [HL] In a genetic cross investigating two unlinked genes, the results are compared to the expected 9:3:3:1 ratio. A chi-squared test yields a value of 6.95. For this test, with 3 degrees of freedom, the critical value at a significance level of p=0.05 is 7.81. Which statement is a valid conclusion from this result?
   Answer: The deviation between observed and expected results is not statistically significant. In a chi-squared test, the null hypothesis (H₀) states that there is no significant difference between observed and expected results. The hypothesis is rejected if the calculated chi-squared value is greater than the critical value. Here, the calculated value (6.95) is less than the critical value (7.81). Therefore, we fail to reject the null hypothesis. This means the observed deviation is not statistically significant and is likely due to random chance.
7. Red-green colour blindness is a human X-linked recessive trait. A female who is a carrier for colour blindness mates with a male who is colour-blind. What is the expected phenotypic ratio among their daughters only?
   Answer: 1 daughter with normal vision : 1 daughter who is colour-blind. Let $X^B$ be the allele for normal vision and $X^b$ be the allele for colour blindness. The carrier female's genotype is $X^B$ $X^b$. The colour-blind male's genotype is $X^b$ Y. The cross is $X^B$ $X^b$ x $X^b$ Y. The possible genotypes for their daughters are $X^B$ $X^b$ (carrier with normal vision) and $X^b$ $X^b$ (colour-blind). These two genotypes have an equal probability of occurring. Therefore, the expected phenotypic ratio among daughters is 1 with normal vision to 1 who is colour-blind.
8. In a plant species, flower colour is determined by a gene with two alleles showing incomplete dominance: R (red) and r (white), with Rr resulting in pink flowers. A plant with pink flowers is self-pollinated. If three offspring are produced from this cross, what is the probability that at least one of them will have pink flowers?
   Answer: 7/8 The cross is Rr x Rr, which produces offspring in the genotypic ratio 1 RR : 2 Rr : 1 rr. The corresponding phenotypic ratio is 1 red : 2 pink : 1 white. The probability of an offspring having pink flowers (Rr) is 2/4 or 1/2. The probability of an offspring not having pink flowers (being RR or rr) is 1 - 1/2 = 1/2. The probability of all three offspring not having pink flowers is (1/2) (1/2) (1/2) = 1/8. The probability of 'at least one' having pink flowers is the complement of 'none have pink flowers', so P(at least one pink) = 1 - P(none pink) = 1 - 1/8 = 7/8.
9. A pedigree analysis for a genetic disorder reveals that the trait appears in every generation. Furthermore, all affected individuals have at least one affected parent, and affected parents can have unaffected offspring. What is the most probable mode of inheritance?
   Answer: Autosomal dominant The trait appearing in every generation strongly suggests a dominant pattern, as recessive traits can skip generations. The fact that affected individuals have an affected parent reinforces this. If it were recessive, two unaffected parents could have an affected child. The fact that affected parents can have unaffected children (e.g., Aa x aa -> aa) is consistent with a dominant trait and rules out some rare forms of inheritance. This pattern is characteristic of autosomal dominant inheritance.
10. A man is a carrier for the autosomal recessive disorder cystic fibrosis. He and his partner, whose family history is unknown, have a child with cystic fibrosis. What must be true about the mother's genotype?
    Answer: She is heterozygous for the allele. Cystic fibrosis is an autosomal recessive disorder. A child with the disorder must have a homozygous recessive genotype (e.g., ff). This child inherited one recessive allele from each parent. We are told the father is a carrier (Ff). Therefore, the mother must also have at least one recessive allele (f) to pass to the child. Since her phenotype is not mentioned as having the disease, it is assumed she is unaffected, meaning she cannot be homozygous recessive (ff). Thus, she must be heterozygous (Ff).
11. Human height shows a wide range of variation and approximates a normal distribution within the population. Which genetic mechanism is the most likely explanation for this observation?
    Answer: Polygenic inheritance where multiple genes have an additive effect. Traits that show continuous variation, such as height, are typically polygenic. This means they are controlled by the cumulative, additive effects of multiple genes. Each gene contributes a small amount to the overall phenotype, leading to a wide spectrum of possible outcomes that often form a bell-shaped (normal) distribution. The other options describe patterns of discontinuous variation.
12. Haemophilia is an X-linked recessive disorder. A phenotypically normal woman whose father had haemophilia marries a phenotypically normal man. What is the probability that their first son will have haemophilia?
    Answer: 1/2 Since the woman's father had haemophilia ($X^h$ Y), he must have passed his $X^h$ allele to her. As she is phenotypically normal, her genotype is $X^H$ $X^h$ (a carrier). The normal man has the genotype $X^H$ Y. The cross is $X^H$ $X^h$ x $X^H$ Y. Their sons can inherit either the mother's $X^H$ (resulting in $X^H$ Y, normal) or her $X^h$ (resulting in $X^h$ Y, haemophilia). Each outcome has an equal chance. Therefore, the probability that a son born to this couple will have haemophilia is 1/2.
13. In certain mice, the allele for a yellow coat (Y) is dominant over the allele for a grey coat (y). However, the homozygous dominant genotype (YY) is lethal during embryonic development. If two yellow mice are crossed, what is the expected phenotypic ratio among their surviving offspring?
    Answer: 2 yellow : 1 grey Since the YY genotype is lethal, any living yellow mouse must be heterozygous (Yy). The cross is therefore Yy x Yy. The initial genotypic ratio of the zygotes is 1 YY : 2 Yy : 1 yy. However, the YY embryos do not survive. Thus, the surviving offspring are only those with genotypes Yy (yellow) and yy (grey). The ratio of these survivors is 2 Yy : 1 yy. This corresponds to a phenotypic ratio of 2 yellow : 1 grey.
14. [HL] Two genes, P and Q, are linked. An individual with genotype PpQq, where alleles P and Q are on one homologous chromosome and p and q are on the other, is test-crossed. Which phenotypic classes among the offspring are evidence that crossing over occurred between the two genes?
    Answer: Only phenotypes corresponding to the genotypes Ppqq and ppQq. The parent's chromosomes have the arrangement (PQ)/(pq). Without crossing over, it produces parental gametes PQ and pq. A test cross would yield offspring with parental phenotypes (P-Q- and ppqq). Crossing over between the genes produces recombinant gametes Pq and pQ. A test cross with these gametes would yield offspring with recombinant phenotypes (P-qq and ppQ-). The presence of these recombinant phenotypes is the evidence that crossing over occurred.
15. [HL] Non-disjunction of the sex chromosomes during meiosis I in a human male results in the formation of abnormal sperm. If one of these abnormal sperm fertilizes a normal egg, which karyotype is a possible outcome for the zygote?
    Answer: XXY In meiosis I, homologous chromosomes separate. Non-disjunction in a male means the X and Y chromosomes fail to separate. This produces two types of abnormal sperm: one containing both X and Y (XY sperm), and one containing no sex chromosome (nullo sperm). A normal egg contains one X chromosome. Fertilization by the XY sperm results in an XXY zygote (Klinefelter syndrome). Fertilization by the nullo sperm results in an XO zygote (Turner syndrome). XYY results from non-disjunction in meiosis II.
16. [HL] In fruit flies, grey body (G) is dominant to ebony body (g), and normal wings (W) are dominant to vestigial wings (w). A fly heterozygous for both traits is test-crossed with a fly that has an ebony body and vestigial wings. If 800 offspring are produced, what is the expected number of flies with a grey body and vestigial wings?
    Answer: 200 This is a dihybrid test cross: GgWw x ggww. The heterozygous parent (GgWw) produces four types of gametes in equal proportion (assuming independent assortment): GW, Gw, gW, and gw. The homozygous recessive parent (ggww) produces only one type of gamete: gw. The resulting offspring genotypes will be GgWw, Ggww, ggWw, and ggww, each with a probability of 1/4. The phenotype 'grey body and vestigial wings' corresponds to the genotype Ggww. Therefore, the expected frequency is 1/4. The expected number of such flies is (1/4) * 800 = 200.
17. A woman with type A blood and a man with type B blood have their first child, who has type O blood. Based on this information, what is the probability that their second child will have type B blood?
    Answer: 1/4 For a child to have type O blood (genotype ii), they must inherit an 'i' allele from each parent. Since the woman has type A blood, her genotype must be $I^A$ i. Since the man has type B blood, his genotype must be $I^B$ i. The cross is $I^A$ i x $I^B$ i. The possible offspring genotypes are $I^A$ $I^B$ (Type AB), $I^A$ i (Type A), $I^B$ i (Type B), and ii (Type O), each with a probability of 1/4. Therefore, the probability of their second child having type B blood ($I^B$ i) is 1/4.
18. [HL] During a dihybrid test cross for two linked genes, an analysis of 800 offspring reveals 128 individuals with recombinant phenotypes. What is the calculated map distance between these two genes?
    Answer: 16.0 cM Map distance is calculated from the recombination frequency. Recombination frequency = (Number of recombinant offspring / Total number of offspring) x 100. In this case, Frequency = (128 / 800) x 100. 128/800 = 0.16. 0.16 x 100 = 16%. One percent recombination frequency is equal to one map unit or centiMorgan (cM). Therefore, the map distance is 16.0 cM.
19. An agricultural scientist has a pea plant that produces yellow seeds, a dominant trait. To determine if this plant is homozygous dominant (YY) or heterozygous (Yy), which cross would provide the most conclusive information in a single generation?
    Answer: Crossing the plant with one that produces green seeds (yy). This scenario requires a test cross. A test cross involves mating an individual with a dominant phenotype but unknown genotype with a homozygous recessive individual. In this case, crossing the yellow-seeded plant with a green-seeded plant (yy) is the test cross. If any offspring produce green seeds, the unknown parent must be heterozygous (Yy). If all offspring produce yellow seeds (after a large sample size), the parent is likely homozygous dominant (YY). This cross is the most direct and informative.
20. The fur colour of Himalayan rabbits is controlled by a gene that codes for a heat-sensitive enzyme required for melanin production. The enzyme is active only at cooler temperatures. A patch of white fur on a rabbit's back is shaved, and an ice pack is secured over the shaved area while the fur regrows. What will be the colour of the newly grown fur in that patch?
    Answer: Black, because the lower temperature from the ice pack allows the enzyme to be active. This is an example of gene expression being influenced by the environment. The genotype for the Himalayan pattern produces a heat-sensitive tyrosinase enzyme. The ice pack creates a local environment of low temperature on the rabbit's back. In this cold spot, the enzyme will be active, leading to the production of melanin. Consequently, the fur that regrows in this specific patch will be black, similar to the fur on the rabbit's naturally cooler extremities.

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